Question #2cc18

1 Answer
Feb 27, 2017

Answer:

#2"HCl"_ ((aq)) + "Ca"("OH")_ (2(s)) -> "CaCl"_ (2(aq)) + 2"H"_ 2"O"_ ((l))#

Explanation:

The idea here is that all the atoms that are present on the reactants' side must also be present on the products' side.

So, you know that the unbalanced chemical equation looks like this

#"HCl"_ ((aq)) + "Ca"("OH")_ (2(s)) -> "CaCl"_ (2(aq)) + "H"_ 2"O"_ ((l))#

It's always a good idea to start with an inventory of what you have in the unbalanced chemical equation

#color(white)(aacolor(blue)(ul(color(black)("the reactants' side")))aaaaaaaaaaaacolor(blue)(ul(color(black)("the products' side")))aaaa#

#color(white)(aaaaaaaaacolor(black)(3 xx "H")aaaaaaaaaaaaaaaaaaaaaaacolor(black)(2 xx "H")aaaaacolor(red)(xx)#

#color(white)(aaaaaaaaacolor(black)(1 xx "Ca")aaaaaaaaaaaaaaaaaaaaaacolor(black)(1 xx "Ca")aaaaacolor(darkgreen)(sqrt())#

#color(white)(aaaaaaaaacolor(black)(1 xx "Cl")aa.aaaaaaaaaaaaaaaaaaaacolor(black)(2 xx "Cl")aaaaacolor(red)(xx)#

#color(white)(aaaaaaaaacolor(black)(2 xx "O")aaaaaaaaaaaaaaaaaaaaaaacolor(black)(1 xx "O")aaaaacolor(red)(xx)#

Start by balancing the chlorine atoms. You have #1# chlorine atom on the reactants' side and #2# on the products' side, so add a coefficient of #2# to hydrochloric acid

#2"HCl"_ ((aq)) + "Ca"("OH")_ (2(s)) -> "CaCl"_ (2(aq)) + "H"_ 2"O"_ ((l))#

You now have

#color(white)(aacolor(blue)(ul(color(black)("the reactants' side")))aaaaaaaaaaaacolor(blue)(ul(color(black)("the products' side")))aaaa#

#color(white)(aaaaaaaaacolor(black)(4 xx "H")aaaaaaaaaaaaaaaaaaaaaaacolor(black)(2 xx "H")aaaaacolor(red)(xx)#

#color(white)(aaaaaaaaacolor(black)(1 xx "Ca")aaaaaaaaaaaaaaaaaaaaaacolor(black)(1 xx "Ca")aaaaacolor(darkgreen)(sqrt())#

#color(white)(aaaaaaaaacolor(black)(2 xx "Cl")aa.aaaaaaaaaaaaaaaaaaaacolor(black)(2 xx "Cl")aaaaacolor(darkgreen)(sqrt())#

#color(white)(aaaaaaaaacolor(black)(2 xx "O")aaaaaaaaaaaaaaaaaaaaaaacolor(black)(1 xx "O")aaaaacolor(red)(xx)#

Now, notice that you can balance the hydrogen atoms and the oxygen atoms by adding a coefficient of #2# to the water

#2"HCl"_ ((aq)) + "Ca"("OH")_ (2(s)) -> "CaCl"_ (2(aq)) + 2"H"_ 2"O"_ ((l))#

You now have

#color(white)(aacolor(blue)(ul(color(black)("the reactants' side")))aaaaaaaaaaaacolor(blue)(ul(color(black)("the products' side")))aaaa#

#color(white)(aaaaaaaaacolor(black)(4 xx "H")aaaaaaaaaaaaaaaaaaaaaaacolor(black)(4 xx "H")aaaaacolor(darkgreen)(sqrt())#

#color(white)(aaaaaaaaacolor(black)(1 xx "Ca")aaaaaaaaaaaaaaaaaaaaaacolor(black)(1 xx "Ca")aaaaacolor(darkgreen)(sqrt())#

#color(white)(aaaaaaaaacolor(black)(2 xx "Cl")aa.aaaaaaaaaaaaaaaaaaaacolor(black)(2 xx "Cl")aaaaacolor(darkgreen)(sqrt())#

#color(white)(aaaaaaaaacolor(black)(2 xx "O")aaaaaaaaaaaaaaaaaaaaaaacolor(black)(2 xx "O")aaaaacolor(darkgreen)(sqrt()))#

And there you have it, the chemical equation is now balanced.